<?php
/**
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 * Copyright © 2018 Alex Yst <mailto:copyright@y.st>
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$xhtml = array(
	'<{title}>' => 'Functions',
	'<{subtitle}>' => 'Written in <span title="College Algebra">MATH 1201</span> by <a href="https://y.st./">Alex Yst</a>, finalised on 2018-02-07',
	'<{copyright year}>' => '2018',
	'takedown' => '2017-11-01',
	'<{body}>' => <<<END
<section id="problem0">
	<h2>Problem 0</h2>
	<p>
		To do this, we need to arbitrarily limit our x value (function input), but we can pull it off.
	</p>
	<p>
		F = {(x,y)|y=(5-x)/x and 0 &lt; x ≤ 5}
	</p>
</section>
<section id="problem1">
	<h2>Problem 1</h2>
	<p>
		The question gives us nothing to go off of.
		We could solve for t, but we don&apos;t have the other half of the equation.
		We could solve for h, but again, we don&apos;t have what lies on the other side of the equals sign.
		The provided equation is identical to that of the difference quotient though, so I think it was intended as such.
		I believe step zero below is the technically-correct answer for what was actually asked of us, as the problem never makes any request for us to find the difference quotient.
		However, if we look for the difference quotient anyway, we can get a much simpler answer that has the same value.
		Had the question asked us for the difference quotient instead of providing the difference quotient&apos;s formula, it&apos;d be much easier to figure out what to do here as well.
	</p>
	<h3>Solution:</h3>
	<p>
		2 - h - 2t, where h ≠ 0
	</p>
	<h3>Steps:</h3>
	<ol start="0">
		<li>
			((2(t + h) - (t + h)<sup>2</sup>) - (2t - t<sup>2</sup>)) / h
		</li>
		<li>
			(((2t + 2h) - (t<sup>2</sup> + h<sup>2</sup> + 2th)) - (2t - t<sup>2</sup>)) / h
		</li>
		<li>
			(2t + 2h - t<sup>2</sup> - h<sup>2</sup> - 2th - 2t + t<sup>2</sup>) / h
		</li>
		<li>
			(2h - h<sup>2</sup> - 2th) / h
		</li>
		<li>
			h * (2 - h - 2t) / h
		</li>
		<li>
			2 - h - 2t
		</li>
	</ol>
</section>
<section id="problem2">
	<h2>Problem 2</h2>
	<p>
		As our textbook states, adding or dividing functions like this requires that x be in the domain of both functions.
		Being in just one is not enough.
		That means that the data points at (4, 7) and (5, 2) are completely irrelevant and must be discarded.
		For both (F+G)(x) and (F/G)(x), the domain is only 1 and three.
		Once we know that, it&apos;s only a matter of adding or dividing the y values.
		In finding (F/G)(3), we get y=5/0.
		We can&apos;t divide by zero, so this combined data point must also be discarded from (F/G)(x).
		(F/G)(x) therefore has only one data point; x can only take on a single value.
		It&apos;s barely a function at all.
	</p>
	<p>
		(F+G)(x) = { (1, 6), (3, 5) }
	</p>
	<p>
		(F/G)(x) = { (1, 0.5) }
	</p>
</section>
END
);
